The concentration, or molality, remains constant. 0000003314 00000 n
If 60.0 mL of 5.0 M HCl is used to make the desired solution, the amount of water needed to properly dilute the solution to the correct molarity and volume can be calculated: In order for the scientist to make 150.0 mL of 2.0 M HCl, he will need 60.0 mL of 5.0 M HCl and 90.0 mL of water. Measurement Units and Concentration Analogies Parts Per Million (ppm) 1 milligram/kilogram (mg/kg) = 1 ppm 1 milligram/liter (mg/l) = 1 ppm 1 microgram/gram (µg/g) = 1 ppm 0.0001 % = 1 ppm 1 ppm Analogies 1 inch in 16 miles 1 minute in two years 1 second in 11.5 days Since volume is subject to variation due to temperature and pressure, molarity also varies by temperature and pressure. c1 and V1 are the concentration and the volume of the starting solution, which is the 5.0 M HCl. <>
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Once we have the mass of acetic acid in kg, we convert from kg to grams: 0.12 kg is equal to 120 g. Next, we use the density of acetic acid (1.05 g/mL at 20 oC) to convert to the requested volume in mL. The SI unit for molar concentration is mol/m3. %����
Next we must divide the moles of N2 by the total number of moles: [latex]x (\text{mole fraction}) = (\frac {\text{ moles } \text{N}_2}{\text{ moles } \text{N}_2 + \text{ moles } O_2})= (\frac {5.6 \text { moles}}{11.9 \text{ moles}})= 0.47[/latex]. xref
1.1 Commonly used concentration units Commonly, concentration units are presented using units in the form of mass per volume (e.g. We also need to convert the the 56.0 mL of water to its equivalent mass in grams by using the known density of water (1.0 g/mL): [latex]56.0\ \text{mL} \times (\frac{1.0 \text{g}}{\text{mL}}) = 56.0\ \text{g}[/latex]. The result is the desired mass of acetic acid that we need to make our 3 m solution: [latex]0.38 \text{ moles KCl} \times (\frac {\text{ kg acetic acid}}{3.0 \text{ moles KCl}}) = 0.12\text{ kg acetic acid}[/latex]. How much water and how much 5.0 M HCl should the scientist use to make 150.0 mL of 2.0 M HCl? endstream
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A solution with a molality of 3 mol/kg is often described as “3 molal” or “3 m.” However, following the SI system of units, mol/kg or a related SI unit is now preferred. Mole fraction describes the number of molecules (or moles) of one component divided by total the number of molecules (or moles) in the mixture. A scientist has a 5.0 M solution of hydrochloric acid (HCl) and his new experiment requires 150.0 mL of 2.0 M HCl. stream
CC licensed content, Specific attribution, http://en.wikipedia.org/wiki/Dilution_(equation), http://en.wikipedia.org/wiki/Molar_concentration, http://en.wiktionary.org/wiki/concentration, http://www.boundless.com//chemistry/definition/intensive-property, http://apchemcyhs.wikispaces.com/Concentration+of+Solutions, http://en.wikipedia.org/wiki/File:SaltInWaterSolutionLiquid.jpg, http://www.youtube.com/watch?v=4VltXjR64SU, http://en.wikipedia.org/wiki/Mole_fraction, http://en.wiktionary.org/wiki/mole_fraction, http://commons.wikimedia.org/wiki/File:Salt_mole_fraction.PNG. Once we know we need to achieve 0.29 moles of BH3, we can use this and the given molarity (3.0 M) to calculate the volume needed to reach 4.0 g. [latex]3.0 \text{ M} = \frac{0.29 \text{moles BH}_3} {\text{V}}[/latex]. Multiplying the mole fraction by 100 gives the mole percentage, also referred as amount/amount percent (abbreviated as n/n%). In chemistry, concentration of a solution is often measured in molarity (M), which is the number of moles of solute per liter of solution. If we have a 1.62 m solution of table sugar (C6H12O6) in water, what is the mole fraction of the table sugar? The molecular weight of urea is 60.16 g/mol and the molecular weight of cinnamic acid is 148.16 g/mol. Concentration Units - Free download as Word Doc (.doc / .docx), PDF File (.pdf), Text File (.txt) or read online for free. What is the mole fraction of nitrogen in the mixture? 0000008678 00000 n
Concentration Units Quantitative study of a solution requires knowing its concentration, the amount of solute present in a given amount of solution. A mixture of known mole fractions can be prepared by weighing the appropriate masses of the constituents. Molality is based on mass, so it can easily be converted into a mass ratio, denoted by w: [latex]\text{bM}_{\text{solute}}=\frac{\text{n}_{\text{solute}}}{\text{m}_{\text{solvent}}} = \frac{\text{w}_{\text{solute}}}{\text{w}_{\text{solvent}}}[/latex]. c2 and V2 are the concentration and the volume of the desired solution, or 150.0 mL of the 2.0 M HCl solution. 0000001336 00000 n
This means that we have 50.0 g of urea and 50.0 g of cinnamic acid. It is not temperature dependent, as opposed to molar concentration, and does not require knowledge of the densities of the phase(s) involved. Mole fraction is the number of molecules of a given component in a mixture divided by the total number of moles in the mixture. If we mass 5.36 g of KCl and dissolve this solid in 56 mL of water, what is the molality of the solution? 0000009298 00000 n
We find that there are 0.138 moles of pentane, 0.116 moles of hexane, and 0.128 moles of benzene. We can easily convert mole percent back to mole fraction by dividing by 100. For example, how much acetic acid, in mL, is needed to make a 3.0 m solution containing 25.0 g of KCN? What is the mole fraction of hexane in this mixture? I. To find the mole fraction, we divide the moles of cinnamic acid by total number of moles: [latex]\text{x} = (\frac {.388 \text{ moles cinnamic acid}}{1.22 \text{ moles solution}})= 0.318[/latex]. Water was added to 25 mL of a stock solution of 5.0 M HBr until the total volume of the solution was 2.5 L. What is the molarity of the new solution? [latex]\text{c}_2 = \frac{(5.0 \text{M})(0.025 \text{L}) }{2.50 \text{L}} =0.05 \text{M}[/latex]. <>>>
For example, diborane (B2H6) is a useful reactant in organic synthesis, but is also highly toxic and flammable. There are 2.6 moles of KCl in a 0.65 M solution that occupies 4.0 L. We can also calculate the volume required to meet a specific mass in grams given the molarity of the solution. endstream
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Therefore: [latex](5.0 \text{ M HCl})(\text{V}_1) = (2.0 \text{ M HCl})(150.0 \text{ mL})[/latex]. [latex]10.0 \text{ grams NaCl} \times \frac{\text{1 mole}}{58.4 \text {g/mole}} = 0.17 \text{ moles NaCl}[/latex]. In chemistry, the mole fraction, xi, is defined as the amount of moles of a constituent, ni, divided by the total amount of moles of all constituents in a mixture, ntot: [latex]\text{x}_{\text{i}}=\frac{\text{n}_{\text{i}}}{\text{n}_{\text{tot}}}[/latex]. 0000002101 00000 n
0
Mole fractions are dimensionless, and the sum of all mole fractions in a given mixture is always equal to 1. 0000000016 00000 n
0000000851 00000 n
We find that the mole fraction of NaCl is 0.0176. 0000007142 00000 n
Molality is a property of a solution that indicates the moles of solute per kilogram of solvent. To calculate the molarity of a solution, the number of moles of solute must be divided by the total liters of solution produced. In some cases, using weight is an advantage because mass does not vary with ambient conditions. Since we are given molality, we can convert it to the equivalent mole fraction, which is already a mass ratio; remember that molality = moles solute/kg solvent. H��UMO�@��W̥���~H�T�8$Te+.���WN��N)���݄O0UQ6Λ7��͌G���x4_���ӳ�1d�3%d�O���dg>yO���g�PJ
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LgÑ�N�Z"�s��1��ev�uz���#���㋩���_^M���]��b However, mol/L is a more common unit for molarity. We will first need to calculate the amount of moles present in 5.36 g of KCl: [latex]\text{ moles KCl} = 5.36 \text{g} \times (\frac{1 \text{ moles}}{74.5 \text{g}}) = 0.0719 \text{ moles KCl}[/latex].